Chứng minh đẳng thức: \(C_n^{k - 1} + C_n^k = C_{n + 1}^k\).

\(C_n^{k - 1} + C_n^k = C_{n + 1}^k\)

\( = \frac{{n!}}{{\left( {k - 1} \right)!.\left[ {n - \left( {k - 1} \right)} \right]!}} + \frac{{n!}}{{k!.\left( {n - k} \right)!}}\)

\( = \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} + \frac{{n!}}{{k!.\left( {n - k} \right)!}}\)

\( = \frac{{n!}}{{k!.\frac{1}{k}.\left( {n - k} \right)!.\left( {n - k + 1} \right)}} + \frac{{n!}}{{k!.\left( {n - k} \right)!}}\)

\( = \frac{{n!.\frac{k}{{n - k + 1}}}}{{k!.\left( {n - k} \right)!}} + \frac{{n!}}{{k!.\left( {n - k} \right)!}}\)

\( = \frac{{n!}}{{k!.\left( {n - k} \right)!}}.\left( {\frac{k}{{n - k + 1}} + 1} \right)\)

\( = \frac{{n!}}{{k!.\left( {n - k} \right)!}}.\left( {\frac{{k + n - k + 1}}{{n - k + 1}}} \right)\)

\( = \frac{{n!}}{{k!.\left( {n - k} \right)!}}.\left( {\frac{{n + 1}}{{n - k + 1}}} \right)\)

\( = \frac{{\left( {n + 1} \right)!}}{{k!.\left( {n - k + 1} \right)!}} = \frac{{\left( {n + 1} \right)!}}{{k!.\left[ {\left( {n + 1} \right) - k} \right]!}}\)

\( = C_{n + 1}^k\)

Vậy \(C_n^{k - 1} + C_n^k = C_{n + 1}^k\).